---
title: Potential Energy
module: Energy
moduleNumber: 3
lessonNumber: 2
order: 302
summary: >
  When a force does the same work no matter which path a particle takes, that
  work can be stored as a function of position alone, and solving for the motion
  becomes bookkeeping instead of integration. We single out the forces that
  qualify — the conservative ones, for which $\oint\vec F\cdot\d\vec r=0$ — define
  their potential energy through $\vec F=-\nabla U$, and use conservation of
  mechanical energy to read speeds, turning points, and equilibria straight off a
  potential curve. Friction breaks the shortcut, so we also track where
  mechanical energy leaks away as heat.
topics: [Energy]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 7 — Conservation of Energy; §§7-1–7-3"
---

## Conservative forces and potential functions

The work done by a conservative force depends only on the initial and final
positions. It is independent of the route between them. Gravity near Earth's
surface and the ideal spring force are standard examples. Kinetic friction is not
conservative: a longer sliding path produces a larger magnitude of friction work,
even when the endpoints are unchanged.

A conservative force $\vec F_c$ defines potential-energy change through

$$
\Delta U=U_f-U_i=-W_c
=-\int_i^f\vec F_c\cdot\d\vec r.
$$

Equivalently, the net work around any closed path is zero:

$$
\oint\vec F_c\cdot\d\vec r=0.
$$

Potential energy belongs to a system, not to one object in isolation. The
gravitational potential energy $mgy$ belongs to the object-Earth system, and
elastic potential energy to the spring-object system. As an object falls, it
gains kinetic energy while the gravitational potential energy of the object-Earth
system drops by the same amount.

> **Definition (potential energy).** A potential-energy function $U(\vec r)$
> is a scalar function whose negative change equals the work of the associated
> conservative force.

Only differences in $U$ appear in force and energy equations. Adding a constant
$C$ to every value, $U'=U+C$, changes no force and no observable energy difference.
The zero may therefore be assigned at any convenient reference configuration. For
near-surface gravity, taking $U=0$ at $y=0$ gives

$$
U_g(y)=mgy,
\qquad
\Delta U_g=mg(y_f-y_i).
$$

The result follows directly from the constant downward force $\vec F_g=-mg\hat y$:

$$
W_g=\int_{y_i}^{y_f}(-mg)\,\d y=-mg(y_f-y_i)=-\Delta U_g.
$$

The approximation $U_g=mgy$ requires nearly constant $g$ over the height change.
It is appropriate for laboratory, projectile, and ordinary engineering scales.

An ideal spring obeying $F_x=-kx$, with $x=0$ at its unstretched length, has

$$
U_s(x)-U_s(0)=-\int_0^x(-kx')\,\d x'
=\frac12kx^2.
$$

Thus $U_s=\tfrac12kx^2$ under the reference $U_s(0)=0$. Compression and extension
of equal magnitude store equal elastic potential energy. The force has opposite
sign on the two sides of equilibrium because it points toward $x=0$.

$$
% caption: The spring force is the negative slope of its parabolic potential.
% Equal extensions on either side of equilibrium have equal potential energy but
% opposite force directions.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.4,0) node[right] {$x$};
  \draw[->, black] (0,0) -- (0,3.6) node[above] {$U$};
  \draw[acc, very thick, domain=0.75:5.25, samples=100]
    plot (\x,{0.05+0.55*(\x-3)*(\x-3)});
  \draw[dashed, black] (3,0) -- (3,3.35);
  \draw[->, black, thick] (1.55,1.55) -- (2.35,1.55) node[midway, above] {$F_x$};
  \draw[->, black, thick] (4.45,1.55) -- (3.65,1.55) node[midway, above] {$F_x$};
  \node[anchor=north] at (3,-0.05) {$0$};
  \node[acc, anchor=west] at (4.75,2.55) {$U_s$};
\end{tikzpicture}
$$

**Force from a potential function.**

In one dimension, differentiating the defining integral gives

$$
F_x=-\frac{\d U}{\d x}.
$$

In three dimensions, the corresponding relation is

$$
\vec F_c=-\nabla U
=-\left(\frac{\partial U}{\partial x}\hat x
+\frac{\partial U}{\partial y}\hat y
+\frac{\partial U}{\partial z}\hat z\right).
$$

The force points in the direction of greatest decrease of potential energy. The
slope, rather than the height of the graph, determines force. A high flat region
has $\d U/\d x=0$ and therefore no $x$-directed force. A steep descending section
has positive $F_x$ because $\d U/\d x<0$.

Equilibrium positions satisfy $F_x=0$, hence $\d U/\d x=0$. The local shape
classifies one-dimensional equilibrium:

- **Stable equilibrium:** $U$ has a local minimum. A small displacement produces a
  restoring force toward the equilibrium point.
- **Unstable equilibrium:** $U$ has a local maximum. A small displacement produces
  a force away from the equilibrium point.
- **Neutral equilibrium:** $U$ is locally constant. Small translations do not
  change the potential energy or produce a restoring force.

The second derivative gives the local test when it is nonzero:

$$
\left.\frac{\d^2U}{\d x^2}\right|_{x_0}>0
\quad\text{stable},
\qquad
\left.\frac{\d^2U}{\d x^2}\right|_{x_0}<0
\quad\text{unstable}.
$$

The spring potential $U=\tfrac12kx^2$ gives $\d^2U/\d x^2=k>0$, so $x=0$ is stable.

**Conservation of mechanical energy.**

The work-kinetic-energy theorem states $\Delta K=W_{\rm net}$. Separate the net
work into conservative and nonconservative contributions:

$$
\Delta K=W_c+W_{\rm nc}=-\Delta U+W_{\rm nc}.
$$

Rearrangement gives the central accounting equation,

$$
\Delta(K+U)=W_{\rm nc}.
$$

When $W_{\rm nc}=0$,

$$
E_{\rm mech}=K+U=\text{constant}.
$$

Mechanical energy is therefore conserved when only conservative forces act. The
conserved quantity is the sum $K+U$; neither piece is conserved alone. A falling
object trades potential energy for kinetic energy, and a projectile rising in
vacuum trades kinetic for potential, each exchange exact.

$$
% caption: With no nonconservative work, energy moves between kinetic and
% gravitational potential forms. The sum remains the same at the high and low
% points of the track.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.6,2.9) .. controls (1.9,2.8) and (2.3,0.95) .. (3.7,0.9)
    .. controls (4.7,0.86) and (5.4,0.88) .. (6.3,0.9);
  \draw[fill=acc!12, draw=acc, thick] (1.2,2.72) circle (2.6pt);
  \draw[fill=acc!12, draw=acc, thick] (5.2,1.0) circle (2.6pt);
  \draw[->, thick] (2.15,2.35) .. controls (2.95,1.4) and (3.55,1.0) .. (4.4,0.95);
  \draw[->, black] (0.55,0.9) -- (0.55,2.72) node[midway, left] {$h$};
  \draw[dashed, black] (0.55,2.72) -- (1.2,2.72);
  \draw[dashed, black] (0.55,0.9) -- (5.2,0.9);
  \node[anchor=south, align=center] at (1.55,2.95) {$K$ small\\$U_g$ large};
  \node[anchor=west, align=center] at (5.5,1.15) {$K$ large\\$U_g$ small};
\end{tikzpicture}
$$

An energy equation requires clear initial and final states. For an object moving
under gravity and an ideal spring,

$$
\frac12mv_i^2+mgy_i+\frac12kx_i^2
=\frac12mv_f^2+mgy_f+\frac12kx_f^2.
$$

Terms absent from a configuration are set to zero only after the reference choices
are stated. For example, a spring term is absent only if no spring is deformed or
if the initial and final spring extensions are both zero; a gravitational term need
not be included when both configurations have equal height.

> **Worked example (spring launch up an incline).** A $0.50\ \mathrm{kg}$ block is launched up a smooth $30^\circ$ incline by a spring of constant $k=400\ \mathrm{N\,m^{-1}}$ compressed $0.20\ \mathrm m$. At the highest point the block is momentarily at rest and the spring is no longer compressed. Taking the launch position as $y_i=0$,
>
> $$
> K_i=0,
> \qquad U_{s,i}=\frac12(400)(0.20)^2=8.0\ \mathrm J,
> \qquad K_f=U_{s,f}=0.
> $$
>
> Conservation of mechanical energy gives
> $$
> 8.0=mg h=(0.50)(9.8)h,
> \qquad h=1.63\ \mathrm m.
> $$
>
> The distance along the incline is $s=h/\sin30^\circ=3.27\ \mathrm m$. The normal force does no work, since it stays perpendicular to the displacement.

## Energy accounting and reference levels

Friction and drag usually make mechanical energy decrease. The equation

$$
\Delta E_{\rm mech}=W_{\rm nc}
$$

is still exact. For kinetic friction of constant magnitude $f_k$ over distance $d$,

$$
W_f=-f_kd.
$$

A system containing the moving object and surface records the mechanical decrease
as an associated thermal-energy increase, approximately

$$
\Delta E_{\rm th}=f_kd
$$

when other transfers are negligible. A complete energy balance includes thermal
energy, chemical energy, internal energy, and energy transferred across the system
boundary by work or heating. Mechanical energy records the kinetic and selected
potential terms within that larger account.

$$
% caption: Friction transfers energy from organized mechanical motion to thermal
% energy in the block-surface system. Mechanical energy decreases, while total
% energy remains conserved for an isolated enlarged system.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.3,0.85) -- (3.05,0.85);
  \foreach \x in {0.35,0.7,1.05,1.4,1.75,2.1,2.45,2.8}
    \draw[black] (\x,0.85) -- (\x+0.22,0.6);
  \draw[thick] (1.0,0.85) rectangle (1.95,1.7);
  \node at (1.47,1.27) {$K+U$};
  \draw[->, black, thick] (1.15,1.9) -- (2.05,1.9) node[right] {motion};
  \node[black, anchor=north] at (1.47,0.5) {mechanical};
  \draw[->, acc, very thick] (3.35,1.25) -- (4.65,1.25) node[midway, above] {friction};
  \draw[black, thick] (4.9,0.7) rectangle (6.6,1.95);
  \node at (5.75,1.32) {$E_{\rm th}$};
  \node[black, anchor=north] at (5.75,0.55) {thermal};
\end{tikzpicture}
$$

External work changes the energy of a selected system. A person pushing a crate is
external to a crate-only system, so the applied-force work is an external transfer.
In a person-crate system, the force is internal and chemical energy decreases as
the crate's kinetic energy and thermal energy change. Both system choices are valid;
mixing their energy terms in one equation is not.

Power provides a rate form of this bookkeeping. If a motor raises a load at constant
speed $v$ vertically, the mechanical output is $P=mgv$. If the motor is not
ideal, its input power exceeds that output. Energy conservation remains exact after
the thermal and other internal transfers of the motor are included.

**Reading a potential-energy diagram.**

A graph of $U(x)$ contains both energetic and dynamical information. For a specified
total mechanical energy $E$, classical motion is possible only where

$$
K=E-U(x)\ge0.
$$

The intersections $E=U(x)$ are turning points: $K=0$ there, so the speed vanishes
before the motion reverses. A horizontal line below a potential barrier confines the
particle to one allowed region. A line above the barrier allows travel between the
regions. The graph describes one coordinate; it does not display time directly, and
equal horizontal distances do not imply equal travel times.

Near a stable equilibrium $x_0$, a smooth potential can be expanded as

$$
U(x)\approx U(x_0)+\frac12\left.\frac{\d^2U}{\d x^2}\right|_{x_0}(x-x_0)^2.
$$

The missing linear term follows from $\d U/\d x=0$ at equilibrium. Defining
$k_{\rm eff}=\d^2U/\d x^2|_{x_0}>0$ gives a local restoring force

$$
F_x\approx-k_{\rm eff}(x-x_0).
$$

Small oscillations near any smooth stable minimum therefore have the same local
form as a spring. The approximation fails when the displacement is large enough for
higher derivatives of $U$ to matter.

> **Worked example (rough incline).** A $4.0\ \mathrm{kg}$ crate moves up a $25^\circ$ incline with initial speed $8.0\ \mathrm{m\,s^{-1}}$. The coefficient of kinetic friction is $0.20$. Both gravity and friction do negative work over the stopping distance $s$:
>
> $$
> 0-\frac12mv_i^2=-mg(s\sin25^\circ)-\mu_kmg\cos25^\circ\,s.
> $$
>
> Cancelling $m$,
> $$
> s=\frac{v_i^2}{2g(\sin25^\circ+0.20\cos25^\circ)}=5.41\ \mathrm m.
> $$
>
> The same result reads $\Delta(K+U_g)=W_f$: gravity enters as a potential change, friction as nonconservative work. Friction cannot be folded into a potential, because its work depends on path length.

**Energy accounting protocol.**

Build an energy equation in this order.

1. State the system and its initial and final configurations.
2. List the energy forms that change within that system.
3. Identify energy transferred across the boundary by external work or heating.
4. Write one equation with the signs fixed by those definitions.

A closed system obeys the broad form

$$
\Delta K+\Delta U_g+\Delta U_s+\Delta E_{\rm th}+\Delta E_{\rm int}=0.
$$

A system with external work $W_{\rm ext}$ and heat transfer $Q$ defined positive
into it has right side $W_{\rm ext}+Q$. Include thermal energy created by
friction when total-energy conservation is claimed. Mechanical-energy conservation is
the restricted case with zero nonconservative work.

Raising the potential reference by $h_0$ adds $mgh_0$ to every gravitational
potential, and both sides of an energy equation gain the same constant, so force
and prediction are untouched. Changing the system boundary is different: it moves
terms between internal energy and external work, and must be applied the same way
throughout one calculation. Evaluate every term between the same two states, so a
valid potential difference is never paired with a friction distance or external-work
interval taken from a different part of the record.

The ledger separates endpoint quantities from path-integrated transfers.

| Quantity | Evaluation | State or path requirement |
| --- | --- | --- |
| Kinetic energy | $K=\tfrac12mv^2$ | speed at each named state |
| Conservative potential | $\Delta U=U_f-U_i$ | same initial and final configuration |
| Nonconservative work | $W_{\rm nc}=\int\vec F_{\rm nc}\cdot\d\vec r$ | stated path interval |
| Thermal change | $\Delta E_{\rm th}$ | included system and measurement window |

**Conservative work as an endpoint calculation.**

For a conservative interaction, the work between two configurations is fixed by the
potential-energy difference alone, with no need to reconstruct the route: every
allowed route between the same endpoints gives the same conservative work. A direct
lift and a long smooth ramp to the same height give equal gravitational work, even
though their applied-force work and travel time differ.

Signs follow from the same relation. When potential energy rises from state i to
state f, conservative work is negative: a particle climbing in gravity or stretching
a spring feels negative conservative work, while an external agent may do positive
work over the same displacement. The work-energy theorem sums all force works, so
keeping the conservative and external terms separate avoids counting one transfer
twice.

> **Worked example (endpoint work).** A $2.0\ \mathrm{kg}$ tool is carried from $1.0$ to $4.5\ \mathrm m$ above a floor by either a ladder or a long ramp. Gravity does $-(2.0)(9.8)(4.5-1.0)=-68.6\ \mathrm J$ on either route. The ramp's extra length adds nothing, since gravity has no component along much of that longer path.

**Reference-level changes leave physics unchanged.**

The zero of a potential-energy function is a reference choice, not a place where the
force vanishes. Replacing $U(y)=mgy$ by $U'(y)=mg(y-y_0)$ shifts every value by the
constant $-mgy_0$; differences are untouched, so conservative work, force, turning
points, and every speed prediction stay the same. The shift must reach every state
in one calculation. Assigning $U=0$ at the floor for the initial state and $U=0$ at
a table for the final state, with no account of the offset, invents an energy
difference that is not there. The same holds for spring energy: any reference works
as long as the constant offset is carried consistently.

Potential diagrams make the shift visible. Adding a constant slides the curve and
any total-energy line up together, and the gap $E-U$, the kinetic energy, is unchanged
only when both move. Sliding the curve while holding a numerical energy line fixed
describes a different state, not a new convention.

> **Worked example (reference shift, box).** A $3.0\ \mathrm{kg}$ box is at height $2.0\ \mathrm m$. With the floor as zero,
> its gravitational potential is $58.8\ \mathrm J$. With a shelf at
> $1.5\ \mathrm m$ as zero, the assigned potential is
> $(3.0)(9.8)(2.0-1.5)=14.7\ \mathrm J$. Raising the box by another
> $0.50\ \mathrm m$ changes either value by $14.7\ \mathrm J$. The two reference
> choices give different labels for the state, not different work by gravity.

## Energy diagrams, forces, and equilibrium

An energy diagram combines a potential curve $U(x)$ with a chosen total mechanical
energy $E$. The vertical gap $E-U(x)$ is kinetic energy. Classical motion is allowed
where that gap is nonnegative and forbidden where the potential lies above the
energy line. A turning point occurs where the gap closes: $E=U(x)$ and the speed is
zero at that instant. The force need not be zero there; a nonzero slope can reverse
the motion immediately after the stop.

Allowed intervals are read horizontally from the intersections of the energy line
and potential curve. A particle in a well with energy below a neighboring barrier
is confined to that well in the one-dimensional classical model. Raising the energy
line past the barrier opens a connecting allowed interval. This is a statement about
available kinetic energy, not a statement that the particle moves at constant speed:
speed is greatest where $U$ is smallest within the allowed region.

The diagram requires one consistent reference. The numerical height of $E$ has no
meaning by itself; only its separation from $U(x)$ matters. At two positions with the
same potential value, the particle has the same kinetic energy and speed magnitude
under mechanical-energy conservation, though its velocity directions may be
opposite on the two sides of a turning point.

$$
% caption: A horizontal total-energy line intersects a one-dimensional potential at
% turning points. Motion is allowed where the line lies above the curve, with speed
% set by the vertical gap; the barrier remains inaccessible until the energy level is
% raised above its peak.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.4,0) node[right] {position};
  \draw[->, black] (0,0) -- (0,4.0) node[above] {energy};
  \draw[very thick, domain=0.6:5.4, samples=100]
    plot (\x,{0.4+0.42*(\x-3)*(\x-3)});
  \draw[dashed, black, thick] (0.4,2.2) -- (5.85,2.2) node[right, black] {$E$};
  \draw[dashed, black] (0.93,0) -- (0.93,2.2);
  \draw[dashed, black] (5.07,0) -- (5.07,2.2);
  \draw[fill=acc, draw=acc] (0.93,2.2) circle (1.6pt);
  \draw[fill=acc, draw=acc] (5.07,2.2) circle (1.6pt);
  \node[anchor=north] at (0.93,-0.05) {turn};
  \node[anchor=north] at (5.07,-0.05) {turn};
  \node[black, anchor=south] at (3,2.28) {gap $=K$};
\end{tikzpicture}
$$

> **Worked example (turning point).** A $0.50\ \mathrm{kg}$ particle moves where the potential is $U(x)=2.0x^2\ \mathrm J$ with $x$ in metres, carrying total mechanical energy $8.0\ \mathrm J$. Turning points satisfy $2.0x^2=8.0$, so $x=\pm2.0\ \mathrm m$. At $x=0$ the potential vanishes and the kinetic energy is $8.0\ \mathrm J$, giving
>
> $$
> v=\sqrt{\frac{2(8.0)}{0.50}}=5.66\ \mathrm{m\,s^{-1}}.
> $$
>
> At either turning point the speed is zero, but the force $F_x=-\d U/\d x=-4.0x$ is nonzero and points back toward the allowed region, so the particle reverses rather than resting.

The energy level and the potential slope answer different questions at the same
position. Reading both prevents a turning point from being treated as an equilibrium.

| Feature of $U(x)$ | Energy statement | Force statement |
| --- | --- | --- |
| $E=U$ | $K=0$ at a turning point | use $F_x=-\d U/\d x$ |
| $\d U/\d x=0$ | no conclusion about $K$ by itself | equilibrium candidate |
| $\d^2U/\d x^2>0$ | local potential minimum | stable small-displacement response |

**Energy-diagram audit.**

Read the diagram in order: fix the coordinate and reference, draw the total-energy
line, mark its intersections with $U(x)$ as candidate turning points, then find the
intervals where $E-U\ge0$. Compute a speed from the vertical gap only after those
steps. Skipping them invites the standard error of reading a potential minimum as a
turning point when the energy line sits well above it. Near a shallow intersection,
the turning coordinate is only as sharp as the measured curve, so quoting it to more
digits than the plot supports overstates the data.

**Force from the slope of a potential.**

The graph height is potential energy; its slope is force, $F_x=-\d U/\d x$. The two
readings are independent: a high flat plateau carries large potential energy and
zero force, while a low but steep region exerts a large force. The negative sign
pushes a particle downhill on the graph, toward lower potential energy. A slope read
from two widely separated points can hide a nearby change in curvature, so close
brackets or a local fit represent the derivative better.

> **Worked example (slope to force).** Near $x=1.2\ \mathrm m$, a measured potential has local slope
> $\d U/\d x=-7.0\ \mathrm{J\,m^{-1}}$. Since one joule per metre is one newton,
> the force component is $F_x=+7.0\ \mathrm N$. At a nearby point where the slope is
> $+3.0\ \mathrm{J\,m^{-1}}$, the force is $-3.0\ \mathrm N$. The reversal follows
> from the slope sign, not from whether the potential value itself is positive.

**Equipotential paths in two dimensions.**

In two or three dimensions, an equipotential curve or surface consists of locations
with the same potential energy. Moving exactly along an equipotential gives
$\Delta U=0$, so a conservative force does zero net work over that displacement.
The force is perpendicular to an equipotential because
$\vec F_c=-\nabla U$ points in the direction of most rapid decrease of the
potential. This is the multidimensional version of the one-dimensional slope rule.

Equipotential geometry separates path length from work. A long path winding around a
contour can have zero conservative work if it stays at the same potential, whereas a
short displacement crossing closely spaced contours can involve a large potential
change and large work magnitude. The density of contour lines indicates the
magnitude of the potential gradient on a map with a stated contour interval. Closely
spaced contours correspond to a larger force magnitude than widely spaced contours
for the same potential spacing.

The force direction is normal to the contour and points toward lower labelled
potential. A particle may still have velocity along a contour, but the conservative
force then has no instantaneous power because it is perpendicular to the velocity.
If another force pushes the particle across contours, the conservative force does
work whose sign follows the direction of the potential change.

$$
% caption: Equipotential contours of a two-dimensional conservative-force field.
% The force is perpendicular to the contour and points toward lower potential;
% displacement along one contour has zero potential change and hence zero
% conservative work, regardless of the distance travelled along it.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (3,1.9) ellipse (0.8 and 0.55);
  \draw[black] (3,1.9) ellipse (1.6 and 1.1);
  \draw[black] (3,1.9) ellipse (2.4 and 1.65);
  \node[black] at (3,1.9) {low $U$};
  \draw[fill=acc!12, draw=acc, thick] (4.6,1.9) circle (2pt);
  \draw[->, acc, thick] (4.6,1.9) -- (3.55,1.9) node[midway, above] {$F$};
  \draw[->, black, thick] (4.6,1.9) -- (4.6,2.8) node[above] {along contour};
  \node[black, anchor=south] at (3,3.62) {higher $U$};
\end{tikzpicture}
$$

**Contour-work interpretation.**

An object is displaced $4.0\ \mathrm m$ along a gravitational equipotential at a
fixed height. Gravity does zero work despite the nonzero distance. A subsequent
$1.5\ \mathrm m$ vertical descent changes potential by
$-mg(1.5\ \mathrm m)$, so gravity does positive work $+mg(1.5\ \mathrm m)$. The
two segments differ because one follows a contour and the other crosses contours.

## Conservative and dissipative energy accounting

An energy balance becomes reliable when conservative potential changes and
nonconservative work are kept in separate columns. For a particle or stated system,
the general mechanical relation is

$$
K_f+U_f=K_i+U_i+W_{\rm nc}.
$$

Here $W_{\rm nc}$ includes work by forces not represented by the selected potential,
such as kinetic friction, drag, or an external drive. A negative nonconservative
work reduces mechanical energy; a positive external push increases it. This equation
does not assert that energy disappears when friction acts. It accounts for the
mechanical subset while leaving thermal or other internal energy changes to the
larger-system accounting.

An energy equation should not absorb friction into a potential simply because the
friction magnitude is constant. Its work depends on the distance travelled and
reverses sign with path direction, so a return path gives additional negative
friction work rather than a recovery of the original mechanical energy. This path
dependence sets friction apart from conservative gravity or an ideal
spring.

> **Worked example (mixed work).** A $2.0\ \mathrm{kg}$ crate starts from rest and descends $1.5\ \mathrm m$ while
> kinetic friction does $-8.0\ \mathrm J$ of work. Gravity decreases potential energy
> by $-(2.0)(9.8)(1.5)=-29.4\ \mathrm J$, so the energy balance gives
>
> $$
> K_f-K_i=-\Delta U+W_f=29.4-8.0=21.4\ \mathrm J.
> $$
>
> The final speed is $\sqrt{2(21.4)/2.0}=4.63\ \mathrm{m\,s^{-1}}$. Gravity supplies
> the conservative transfer; friction reduces the mechanical-energy gain. Treating
> friction as a positive potential change would reverse its physical effect.

**Accounting decision procedure.**

Sort the forces first: near-surface gravity and an ideal spring supply potential
differences, while kinetic friction, drag, and a prescribed drive stay as work terms
unless the system is enlarged to absorb their internal-energy changes. Fix the
boundary before substituting, and let each transfer appear once. If gravity enters
as $\Delta U_g$, it does not also enter as gravitational work; if friction enters as
$W_f$, there is no second “friction potential.” External work carries its own sign
from the force and its point's displacement, positive for a lifting motor and
negative for a brake on the moving object.

> **Worked example (mixed accounting check).** An object rises while a motor does $+50\ \mathrm J$ of work, gravity potential
> increases by $30\ \mathrm J$, and drag does $-8\ \mathrm J$ of work. The kinetic-
> energy change is $50-30-8=+12\ \mathrm J$. The calculation has one motor-work term,
> one gravitational-potential term, and one drag-work term. Replacing the gravity term
> with both $+30\ \mathrm J$ potential change and a separate $-30\ \mathrm J$
> gravitational-work term would count the same conservative interaction twice.

[^tipler7]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §§7-1–7-3.

**Quantitative potential-reference shifts.**

Evaluate the same two configurations under two zeros to watch the cancellation
directly. Floor-zero and table-zero potentials differ by a constant at every height,
and the numerical total energy shifts by that same constant when quoted in the new
convention, so differences, conservative work, and speeds are unchanged.

> **Worked example (reference shift, gravity).** A $2.0\ \mathrm{kg}$ object moves from $y=1.0$ to $y=3.5\ \mathrm m$. With floor
> zero, the potential change is $(2.0)(9.8)(2.5)=49\ \mathrm J$. If the reference is
> raised to $y_0=2.0\ \mathrm m$, the two potentials are
> $U_i'=-19.6\ \mathrm J$ and $U_f'=29.4\ \mathrm J$, whose difference is still
> $49\ \mathrm J$. Gravity does $-49\ \mathrm J$ in either convention.

**Force-potential consistency checks.**

A proposed potential function and force law must agree through the negative-gradient
relation. In one dimension, differentiating the potential gives the predicted force;
integrating the negative force gives potential differences up to an arbitrary
constant. These two operations are inverse checks. If a stated force is not the
negative slope of the stated potential, the pair cannot describe the same
conservative interaction over the interval.

$$
% caption: A potential curve and its force graph must be related by negative slope.
% The parabolic potential has a linear restoring force crossing zero at the same
% equilibrium point; the opposite signs on either side of the minimum give the
% force direction toward lower potential.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->, black] (0,0) -- (4.4,0) node[right] {position};
    \draw[->, black] (0,0) -- (0,3.3) node[above] {$U$};
    \draw[very thick, domain=0.5:3.5, samples=100]
      plot (\x,{0.2+0.95*(\x-2)*(\x-2)});
    \draw[dashed, black] (2,0) -- (2,0.2);
    \node[anchor=north] at (2,-0.05) {$x_0$};
  \end{scope}
  \begin{scope}[shift={(5.9,1.5)}]
    \draw[->, black] (0,-1.5) -- (0,1.75) node[above] {$F_x$};
    \draw[->, black] (0,0) -- (4.4,0) node[right] {position};
    \draw[very thick] (0.5,1.4) -- (2,0) -- (3.5,-1.4);
    \draw[fill=acc, draw=acc] (2,0) circle (1.6pt);
    \draw[dashed, black] (2,-1.5) -- (2,0);
    \node[anchor=north] at (2,-1.55) {$x_0$};
  \end{scope}
\end{tikzpicture}
$$

> **Worked example (force-law consistency).** For $U(x)=3.0x^2-4.0x\ \mathrm J$, the force is
> $F_x=-(6.0x-4.0)=4.0-6.0x\ \mathrm N$. At $x=0$, force is positive; at
> $x=1.0\ \mathrm m$, force is negative. The equilibrium is at
> $x=4.0/6.0=0.667\ \mathrm m$, and the positive second derivative confirms a stable
> minimum. A force law $6.0x-4.0$ would have the opposite direction and would not
> match this potential.

**Full energy solution with a dissipative segment.**

A motion can contain conservative and dissipative stages within one energy method.
Locate the interval where each nonconservative force acts and calculate its work only
over that portion of the path. For a block launched by a spring,
climbing smoothly, and then crossing a rough section, elastic and gravitational
potential changes are conservative terms. Kinetic friction on the rough segment is
a negative work term. The final kinetic energy follows from one complete account or
from sequential stage balances that use the same system and reference choices.

The route must be physically feasible at every stage. If the available mechanical
energy is exhausted on the smooth rise, the block never reaches the rough segment and
its friction work is zero. If it reaches the rough segment but stops before its end,
integrate friction only through the actual stopping distance. A negative final
kinetic energy is not an energy prediction; it is evidence that the assumed final
configuration is unreachable under the stated model.

$$
% caption: A spring launch, smooth climb, and rough final segment require one
% energy account with distinct terms. Spring and gravitational potential changes are
% endpoint contributions; kinetic friction removes mechanical energy only on the
% labelled rough interval and determines whether the stated final point is reachable.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.3,0.6) -- (1.9,0.6) -- (3.9,2.3) -- (6.3,2.3);
  \foreach \x in {4.4,4.7,5.0,5.3,5.6,5.9}
    \draw[black] (\x,2.3) -- (\x+0.22,2.55);
  \draw[black, thick] (0.32,0.6) -- (0.32,1.25);
  \draw[black] (0.35,0.9) -- (0.5,1.08) -- (0.65,0.72) -- (0.8,1.08)
    -- (0.95,0.72) -- (1.1,1.08) -- (1.2,0.9);
  \draw[fill=acc!12, draw=acc, thick] (1.2,0.6) rectangle (1.78,1.18);
  \draw[->, acc, thick] (2.25,1.0) -- (3.35,1.95) node[midway, above, sloped] {smooth};
  \node[black, anchor=north] at (0.8,0.55) {spring};
  \node[black, anchor=south] at (5.3,2.62) {rough};
\end{tikzpicture}
$$

> **Worked example (dissipative stage).** A $0.50\ \mathrm{kg}$ block receives $18\ \mathrm J$ from a compressed spring,
> then climbs smoothly through $2.0\ \mathrm m$ before entering a rough horizontal
> surface. The climb raises gravitational potential by $(0.50)(9.8)(2.0)=9.8\ \mathrm
> J$, leaving $8.2\ \mathrm J$ of kinetic energy at the rough section. Kinetic
> friction there has magnitude $2.0\ \mathrm N$. The block can travel
> $8.2/2.0=4.1\ \mathrm m$ before stopping. Any claimed final location beyond
> $4.1\ \mathrm m$ on that rough surface would require negative kinetic energy.

**Consistency checks across the three descriptions.**

Potential, force, and energy diagrams describe one model and must agree at every
marked point: a stable equilibrium is a local minimum with zero force that confines
a low enough energy line between turning points, and an unstable equilibrium is a
local maximum with zero force and a barrier in place of a well. Two numerical checks
back this up. Recomputing a potential difference under two zeros must return the same
conservative work; a speed that shifts when only the reference zero moves means a
constant was applied to one state and not the other. A friction stopping distance
must land inside the rough interval it was derived for, or the final state moves to
that boundary and the leftover kinetic energy is reported.

> **Worked example (effective-potential threshold).** Suppose an effective-potential well has a minimum at $4\ \mathrm J$ and an adjacent
> barrier at $13\ \mathrm J$. A particle with total energy $10\ \mathrm J$ can move
> within the well but cannot cross the barrier; it has at most $6\ \mathrm J$ of
> kinetic energy at the minimum. Raising the energy to $13\ \mathrm J$ reaches the
> barrier top with zero kinetic energy. Values above $13\ \mathrm J$ permit passage
> in the one-dimensional conservative model, provided no dissipative stage removes
> the required energy first.

## Effective potentials and multistage motion

When several conservative interactions act along one coordinate, their potential
functions can be added to form an effective potential. The resulting curve contains
the combined force information:

$$
F_x=-\frac{\d U_{\rm eff}}{\d x}.
$$

For example, a vertical spring with a hanging mass has an elastic contribution and
a gravitational contribution. A constant external conservative force can likewise
add a linear term. The equilibrium position is found from the zero slope of the
sum, not by separately setting each individual force to zero. Individual forces can
be large at equilibrium while their vector sum is zero.

A local minimum of the effective potential is stable because a small displacement
raises the potential and the negative slope produces a restoring force. A local
maximum is unstable because a small displacement lowers the potential on either side
and the force drives the object farther away. A flat region has zero force over the
displayed interval and is neutral only within the model's resolution. Stability is
therefore a local curvature statement, not a claim that the potential value is small
or negative.

The total-energy line gives a further reading. A particle with energy slightly above
a stable minimum is confined between nearby turning points and oscillates in the
well. Raising the energy above an adjacent maximum opens an escape route. The
barrier height is measured relative to the well minimum, so a vertical reference
shift changes neither stability nor the threshold energy difference.

$$
% caption: An effective potential combines conservative contributions into one
% curve. The left minimum is stable and confines low-energy motion between turning
% points, while the central maximum is unstable and becomes passable only when the
% total-energy line rises above the barrier.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.3,0) node[right] {coordinate};
  \draw[->, black] (0,0) -- (0,3.4) node[above] {energy};
  \draw[very thick, domain=0.7:4.6, samples=120]
    plot (\x,{5.5-1.2*((\x*\x*\x)/3 - 2.6*\x*\x + 5.76*\x)});
  \draw[dashed, black, thick] (0.35,1.6) -- (4.85,1.6) node[right, black] {$E$};
  \draw[fill=acc, draw=acc] (1.6,0.79) circle (1.6pt);
  \draw[fill=acc, draw=acc] (3.6,2.39) circle (1.6pt);
  \node[black, anchor=north] at (1.6,0.68) {stable};
  \node[black, anchor=south] at (3.6,2.5) {unstable};
\end{tikzpicture}
$$

**Combined-potential equilibrium.**

A vertical spring with upward coordinate $x$ has
$U_{\rm eff}=\tfrac12kx^2+mgx$. The equilibrium condition is
$kx+mg=0$, so $x_0=-mg/k$. The spring is stretched downward by this amount. Since
$\d^2U_{\rm eff}/\d x^2=k>0$, the equilibrium is stable. Adding a constant to this
potential changes neither $x_0$ nor the restoring-force behaviour.

**Conservation through a multi-stage motion.**

Split a motion at each point where the force model or configuration changes: a spring
acts during launch and then releases, a block runs smooth then rough, a motor cuts
off at a marked position. Each stage inherits the exit state of the one before, and a
force contributes only on the stage where it acts, with zero work elsewhere. That
keeps a spring force from being carried past release and rough-surface friction from
being applied to a smooth section. A single equation covers the whole motion only if
every potential and work term is scoped to its interval.

A conservative stage passes mechanical energy through unchanged; a friction stage
subtracts the friction work over its own path length. Because kinetic energy vanishes
at a turning point, the current potential plus the accumulated nonconservative work
fixes whether the particle reverses before reaching a later stage. A route can be
energetically closed even when each local equation is correct, if the mechanical
energy runs out before the next region begins.

> **Worked example (three-stage motion).** A $1.0\ \mathrm{kg}$ block leaves a compressed spring with $30\ \mathrm J$ of mechanical energy. It rises smoothly by $2.0\ \mathrm m$, raising gravitational potential by $19.6\ \mathrm J$ and leaving $10.4\ \mathrm J$ of kinetic energy. It then crosses a rough horizontal section where friction does $-6.0\ \mathrm J$ of work, so the block leaves with $4.4\ \mathrm J$ of kinetic energy, a speed $\sqrt{2(4.4)/1.0}=3.0\ \mathrm{m\,s^{-1}}$. The spring term belongs to the launch stage alone and is dropped once contact ends.

**Path-dependent work versus potential difference.**

Potential difference is an endpoint quantity for a conservative interaction. Work
by kinetic friction is a path quantity because it depends on the distance travelled
along the contact. Two routes with the same initial and final heights therefore have
the same gravitational potential change but can have different friction work if
their lengths differ. The resulting final kinetic energies need not match, even
though the conservative contribution is identical.

Comparing the two paths determines whether a potential-energy shortcut is
valid. If the force can be written as the negative gradient of a single-valued
potential over the region, use its endpoint difference. If its direction or
magnitude depends on the route, speed, or history, retain it as a work term. Kinetic
friction, drag, and applied forces with prescribed operating intervals belong in the
second category. A constant force can be conservative, but a force of constant
magnitude that follows the direction opposite motion is not.

A closed route provides a simple test. Conservative work sums to zero after return
to the starting configuration. Friction work remains negative for each sliding leg,
so it continues to reduce mechanical energy around the loop. This difference is not
an optional convention; it determines whether one energy value can describe all
routes to the same endpoint.

$$
% caption: Two paths reach the same final height and therefore have the same
% gravitational potential change. The longer rough path has a larger negative
% friction-work contribution, so its final kinetic energy is smaller even though
% both routes share identical conservative endpoint data.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (0.3,0.5) -- (6.3,0.5);
  \draw[fill=white, draw=black, thick] (1.0,0.9) circle (2.2pt);
  \draw[fill=white, draw=black, thick] (5.3,2.7) circle (2.2pt);
  \node[black, anchor=east] at (0.85,1.05) {start};
  \node[black, anchor=west] at (5.5,2.7) {same height};
  \draw[->, acc, thick] (1.15,1.0) -- (5.15,2.6) node[midway, below, sloped] {smooth, direct};
  \draw[->, black, thick] (1.05,1.05) .. controls (1.9,3.45) and (4.3,3.45) .. (5.2,2.75);
  \node[black, anchor=south] at (3.05,3.3) {rough, longer};
  \draw[dashed, black] (1.0,2.7) -- (5.3,2.7);
  \draw[dashed, black] (1.0,0.9) -- (1.0,2.7);
\end{tikzpicture}
$$

> **Worked example (route comparison).** Two $1.0\ \mathrm{kg}$ carts rise through the same $1.2\ \mathrm m$ height from
> the same initial speed. Cart A follows a smooth route. Cart B follows a rough route
> of length $5.0\ \mathrm m$ with kinetic-friction magnitude $2.0\ \mathrm N$.
> Both have gravitational potential increase $+11.8\ \mathrm J$, but Cart B also has
> friction work $-10.0\ \mathrm J$. Its kinetic energy at the final height is
> $10.0\ \mathrm J$ lower than Cart A's. No single friction potential difference can
> reproduce that route-length dependence.

**Path-dependence audit.**

Before trading a work term for a potential difference, compare two routes between the
same endpoints. If the contribution changes with the route's length, shape, or
direction, it stays a path-work term, while the conservative part is still captured
exactly by its endpoint difference. Over a closed cycle the split is stark: gravity
and an ideal spring return what they took, but kinetic friction drains energy on
every leg, so a particle can come home to the same potential with less kinetic energy
than it started with.

**Active-interval scoping.**

A work term is bound to where its force acts: friction only while the surfaces slide,
a spring only while it is deformed, an external drive or brake only over its stretch
of the path. Noting each interval beside the energy equation keeps a correct
potential curve from being paired with a dissipation or drive model that does not
apply there.